Expected number of empty cells

Question

5 balls are distributed independently and completely at random into 5 cells. What is the expected number of empty cells?

How is this question solved using indicator variables? What I have so far is that with $A_i$ = the no. of balls in the ith cell, $E(A_i) = 1/5$. This leads to $E(A) = 1$ but it seems wrong. Can anyone explain how this is right? If it's wrong than what is wrong in my thinking?

https://math.stackexchange.com/questions/28930/another-balls-and-bins-question — StubbornAtom, Dec 27 '19 at 08:13

score 1 · Accepted Answer · answered Oct 04 '17 at 21:25

1

Let $A_k = \{\text{k cells empty}\}$

Then the expected number of empty cells is $$ E \left(1_{A_1} + 2\;1_{A_2} + 3\;1_{A_3} + 4\;1_{A_4} \right) = P(A_1) + 2P(A_2) + 3P(A_3) + 4P(A_4). $$

Can you take it from here?

answered Oct 04 '17 at 21:25

Epiousios

3,194

I didn't know indicator variables could be used this way. Thanks – user478136 Oct 04 '17 at 21:31

Expected number of empty cells

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